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Outer Automorphism Groups of Bieberbach Groups

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Outer Automorphism Groups of Bieberbach Groups Outer Automorphism Groups of Bieberbach Groups Andrzej Szczepa´nski ∗ Abstract Let Γ be a Bieberbach group it is a fundamental group of a flat manifold. In this paper we give necessary and sufficient conditions on Out(Γ) to be infinite. We also compare our result with the previous one ([8], Theorem 7.1) about Anosov automorphisms of flat manifolds. We give several examples of Bieberbach groups. 1 Introduction Let Γ be a Bieberbach group. By definition it is the torsion - free group given by a short exact sequence (∗)1! Zn ! Γ ! G ! 1; where G is a finite group (called the holonomy group of Γ) and Zn is a free abelian group of rank n which is a maximal abelian subgroup of Γ. Let T : G ! GL(n; Z) be the representation which the group Γ defines on Zn by conjugations. The rank of Γ is n: It is well known that Γ is isomorphic to a uniform discrete subgroup of the group E(n) of the rigid motions of the n - dimensional Euclidean space Rn. The orbit space Rn=Γ=M is a closed flat manifold, for short a flat manifold, with fundamental group Γ. In this note we are going to give necessary and sufficient conditions for Out(Γ) = Aut(Γ) / Inn(Γ) to be infinite. ∗This work was supported by Polish grant KBN nr. 0627/P3/93/04. Received by the editors October 1995. Communicated by A. Warrinier. 1991 Mathematics Subject Classification : Primary 20E36, 20F34, 57R50, 58D05. Bull. Belg. Math. Soc. 3 (1996), 585{593 586 A. Szczepa´nski The group Out(Γ) is isomorphic to Aff(M)=Af f0(M), where Aff(M) denotes the group of affine automorphisms of M and Aff0(M) denotes the identity compo- 1 β1 nent of Aff(M) which is isomorphic to (S ) ,whereβ1 is the first Betti number of M (see [2] p. 214). If Γ has trivial centre, then Out(Γ) gives us complete information about exten- sions of Γ by integers. Using this information and the Calabi reduction theorem one can study the problem of the classification of flat manifolds. (See [10], [11].) In this paper we shall prove: Theorem A. The following conditions are equivalent: (i) Out(Γ) is infinite. (ii) The Q-decomposition of T has at least two components of the same isomor- phism type or there is a Q-irreducible component which is reducible over R. The main inspiration for us is [8], where flat manifolds admitting Anosov diffeo- morphisms are characterized; the main result of that paper can be stated as: Theorem B. ([8] Theorem 6.1) The following conditions for a flat manifold M are equivalent: (i) M supports an Anosov diffeomorphism. (ii) Each Q-irreducible component of T which occurs with multiplicity one is reducible over R. We illustrate the links between these results by a series of examples of Bieberbach groups. (See Section 3.) We also give an estimate for the magnitude of Out(Γ). In particular we shall prove: Proposition. For every n 2 N and every l = 1,2,...,n-1 there is a Bieberbach group Γl of rank n such that Out(Γl) is quasi - isometric to GL(l,Z). We recall the definition of quasi - isometry in Section 3. The properties of the group Out(Γ) were studied in [3] and in Chapter 5 of [2]. We assume familiarity with those papers and freely use the methods described there. It was pointed out to us by Wilhelm Plesken that Theorem A can be also proved by a different method (see Theorem 3.41 of [1]). Acknowledgement. This article was written while I was a visitor at E.N.S. Lyon; I would like to thank the director and the staff for their hospitality. I am also grateful to Christophe Bavard and Boris Venkov for helpful conversations and referee for his (her) valuable suggestions. I also thank the European Community - fellowship for financial support. Outer Automorphism Groups of Bieberbach Groups 587 2 Proof of Theorem A We start with the observation that the group Out(Γ) is infinite if and only if the centralizer CGL(n;Z)(T (G)) of the subgroup T(G) in GL(n,Z) is infinite. That is an easy exercise from Chapter 5 of [2]. We need the lemma: Lemma 1. Let S be a rational representation of a finite group of dimension m. ThenScommuteswithanelementK 2 GL(m; Z) of infinite order if and only if S commutes with an element K1 2 GL(m; Q) of infinite order whose characteristic polynomial has integer coefficients with a unit constant term. ProofofLemma1. If K exists, it will do for K1.IfK1 exists, its rational 2 −1 2 canonical form R GL(m; Z). R = P1 K1P1 for some P1 GL(m; Q). Let h be −1 the product of the denominators of the elements in P1 and P1 . Then there is an integer l ≥ 1 such that h divides all the entries of Rl − I -itis ! l 2 enough to consider the map q : GL(m; Z) GL(m; Zh). Then K1 GL(m; Z) will do for K. ProofofTheoremA.(i) ) (ii). If Out(Γ) is infinite then CGL(n;Z)(T (G)) is n infinite. We now change the basis of Q so that T splits up as T1 ⊕ T2 ⊕ ::: ⊕ Tk with Ti Q-irreducible for each i = 1,2,...,k. ∼ We assume that Ti =6 Tj for i =6 j and all the Ti are R-irreducible. We shall prove that in this case the group CGL(n;Z)(T (G)) is finite. Suppose H 2 CGL(n;Z)(T (G). 0 The new matrix H = P −1HP,whereP is the matrix of the new basis, will have a single block corresponding to each Ti and commuting with it. The characteristic polynomial of this block will divide the characteristic polynomial of H,sothisblock will do for K1 in Lemma 1. Hence, it is enough to prove our statement for T Q-irreducible and R-irreducible. We distinguish 3 cases. 1) T is absolutely irreducible. Then, by Schur's lemma, the only matrices com- muting with T are scalar matrices. But the determinant of the matrix is +1 or -1 and a matrix comes from GL(n,Z). Hence scalars are roots of the unity and the matrix has finite order. 2) T decomposes over C with Schur index one. Choosing a suitable basis for Kn, where K is a minimal splitting field for T , the matrix of T will be ! T1 0 ¯ , 0 T1 ¯ with T1 absolutely irreducible and as Schur index is one, T1 is not equivalent to T1. Again by Schur's lemma the only commuting matrices which come from GL(n,Q) are of the form ! λI 0 , 0 λI¯ with λ 2 K. If the characteristic polynomial is in Z[x] with unit constant term then j λλ¯ j = 1. Hence λ is on the unit circle. But λ 2 K, a complex quadratic extension 588 A. Szczepa´nski of Q, and so, by Dirichlet's unit theorem, λ is a root of unity. Hence our matrix has finite order. 3) T decomposes over C with Schur index two. Let K be as in the case 2. Choosing a suitable basis for Kn, the matrix of T will be ! T1 0 ¯ , 0 T1 ¯ with T1 absolutely irreducible and T1 ∼ T1.LetJ 2 GL(n=2;K) be such that ¯ ¯ ¯ JT1 = T1J.ThensinceJJ commutes with T1, JJ = κI with κ 2 Q = K \ R.If κ>0then ! 0 J¯ , J 0 whose square is κI has real eigenvalues and commutes with T , which is irreducible over R. As it is not a multiple of the identity, this contradicts Schur's lemma, and so κ<0. Now, by Schur's lemma again, every matrix which commuts with T and comes from GL(n,Q)mustbeoftheform ! λI νJ¯ . νJ¯ λI¯ Note that the determinant of this matrix is (λλ¯ − κνν¯)n=2. If its characteristic polynomial is in Z[x] with unit constant term then λλ¯ − κνν¯ = 1. It cannot be -1 as it is positive. Then j λ |≤ 1, so j λ + λ¯ |≤ 2. The characteristic polynomial of the above matrix is (x2 − (λ + λ¯)x +1)n=2. Since the zeros of x2 + ax + 1 are all roots of unity for a =0; −1; 1; −2; 2, we can assume that (λ + λ¯)=2. Hence λ = 1. Because any power of the above matrix has the same properties of the characteristic polynomial and eigenvalues, the only possibility is ν = 0. Hence, again our matrix has finite order. Now we can start the proof of the second part of the Theorem (ii)) (i). If T has at least two components of the same isomorphism type it is easy to construct an integral matrix H1 of infinite order which commutes with T (see [8] p.311) as 0 −1 0 Q-representation. Then if we write H1 = PH1P then H1 satisfies the conditions of Lemma 1. So, assume that T has one component which is R-reducible. Let K be a minimal splitting field whose intersection with R is non-trivial. We may assume that K is a subfield of Q(ζ), where ζ is a primitive j G jth root of unity, and so Γ(K=Q)is abelian. K is not a complex quadratic extension of Q, and so, by Dirichlet's unit theorem, there are in K algebraic units none of whose conjugates (in the Galois sense) are on the unit circle. σi Let λ be such a unit and let fσigi be the Galois group Γ(K=Q).
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